QCE Mathematical Methods - Unit 3 - Discrete random variables
Bernoulli Distributions | QCE Mathematical Methods
Learn Bernoulli distributions for QCE Mathematical Methods, including two-outcome modelling, parameter p, mean and variance.
Updated 2026-05-18 - 5 min read
QCAA official coverage - Mathematical Methods 2025 v1.3
Exact syllabus points covered
- Use a Bernoulli random variable as a model for two-outcome situations.
- Identify contexts suitable for modelling by Bernoulli random variables.
- Recognise and determine the mean $p$ and variance $p(1-p)$ of the Bernoulli distribution with parameter $p$.
- Model and solve problems that involve Bernoulli random variables and associated probabilities, with and without technology.
A Bernoulli random variable models one trial with exactly two outcomes. We usually call them success and failure, but those words are labels, not value judgments. A "success" just means the outcome we are counting.
For a Bernoulli random variable:
$ X= \begin{cases} 1, & \text{success}\\ 0, & \text{failure} \end{cases} $
If the probability of success is $p$, then:
$ P(X=1)=p $
and:
$ P(X=0)=1-p $
Original Sylligence diagram for bernoulli binomial flow.
That is the whole distribution. A Bernoulli variable does not need a long table, but writing it as a table is useful when you are checking expected value or variance:
| $x$ | $0$ | $1$ | | --- | --- | --- | | $P(X=x)$ | $1-p$ | $p$ |
The values $0$ and $1$ are chosen deliberately. They let the random variable behave like an indicator: $1$ means the event happened, $0$ means it did not. That is why Bernoulli variables are so useful in modelling, simulations and later sampling-proportion work.
When to use Bernoulli
Use a Bernoulli model when:
- there is one trial
- there are only two outcomes
- you can clearly define the success outcome
- the probability of success is $p$
Examples include one coin toss, one product being defective or not defective, one survey respondent answering yes or no, or one free throw being made or missed.
The word "success" can be chosen either way. If a question asks about a defective product, it is often easiest to define success as "defective" because that is the outcome being counted. If the question asks about a working product, define success as "working" instead. The mathematics is the same; only the value of $p$ changes.
Common modelling contexts
| Situation | Reason it is Bernoulli | Possible $p$ | | --- | --- | --- | | One fair coin is tossed | heads or not heads | $0.5$ | | One die is rolled and success is "rolling a $6
quot; | $6$ or not $6$ | $\frac{1}{6}$ | | One multiple-choice answer is guessed from four options | correct or incorrect | $\frac{1}{4}$ | | One manufactured item is inspected | defective or not defective | defect probability |Notice that the original situation may have more than two raw outcomes. A die has six faces, but if success is "rolling a $6
quot;, the model collapses the experiment into two categories: $6$ and not $6$.Mean and variance
For $X\sim \operatorname{Bernoulli}(p)$:
$ E(X)=p $
$ \operatorname{Var}(X)=p(1-p) $
The mean makes sense because after many trials, the average number of successes per trial approaches the probability of success.
You can also derive it straight from the probability table:
$ E(X)=0(1-p)+1(p)=p $
For variance, use either definition. The shortcut is quick because $X^2=X$ when $X$ is only $0$ or $1$:
$ E(X^2)=0^2(1-p)+1^2(p)=p $
$ \operatorname{Var}(X)=E(X^2)-[E(X)]^2=p-p^2=p(1-p) $
This derivation is worth knowing because it explains where the formula comes from. It is not just something to memorise.
Reading the parameter
If $X\sim \operatorname{Bernoulli}(0.37)$, then $p=0.37$. That means:
$ P(X=1)=0.37 $
and:
$ P(X=0)=0.63 $
The distribution is fully determined by $p$. There is no separate value for $n$ because there is only one trial.
Worked example
Interpreting p
$p$ is a population or model probability. It is not the same as an observed result. If a basketball player has a free throw probability of $0.80$, they can still miss the next shot. Probability describes long-run behaviour, not a guarantee on a single trial.
Bernoulli in simulations
Technology often simulates Bernoulli trials by generating a random number between $0$ and $1$. For example, to model $p=0.72$, treat the trial as a success if the random number is less than $0.72$ and a failure otherwise. Repeating that many times should give a sample mean close to $0.72$, but not exactly equal every time.
This is the bridge to binomial distributions: one Bernoulli result is a $0$ or $1$, while the sum of many independent Bernoulli results is the number of successes.